$\overline{AC}$ is $9$ units long $\overline{BC}$ is $3$ units long $\overline{AB}$ is $3\sqrt{10}$ units long What is $\tan(\angle BAC)$ ? $A$ $C$ $B$ $9$ $3$ $3\sqrt{10}$
Explanation: SOH CAH TOA an = pposite over djacent opposite $= \overline{BC} = 3$ adjacent $= \overline{AC} = 9$ $\tan(\angle BAC)=\dfrac{3}{9}$ $=\dfrac{1}{3}$